The first term of a geometric series is 20 and the common ratio is \( \frac{7}{8} \) The sum to infinity of the series is \( S_\infty \) (a) Find the value of \( S_\infty \) The sum to \( N \) terms of the series is \( S_N \) (b) Find, to 1 decimal place, the value of \( S_{12} \) (c) Find the smallest value of \( N \), for which \( S_\infty - S_N < 0.5 \) - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 1

The sum of the first ( N ) terms of a geometric series is given by:

$S_N = a \frac{1 - r^N}{1 - r}$

For ( S_{12} ), we substitute ( a = 20 ), ( r = \frac{7}{8} ), and ( N = 12 ):

$S_{12} = 20 \frac{1 - \left(\frac{7}{8}\right)^{12}}{1 - \frac{7}{8}} = 20 \frac{1 - \left(\frac{7}{8}\right)^{12}}{\frac{1}{8}} = 160 \left(1 - \left(\frac{7}{8}\right)^{12}\right)$

Calculating ( \left(\frac{7}{8}\right)^{12} ) gives approximately 0.216. Thus:

$S_{12} \approx 160 \left(1 - 0.216\right) = 160 \times 0.784 = 125.44$

Rounding to 1 decimal place, ( S_{12} \approx 125.4 ).

We need to solve:

$S_\infty - S_N < 0.5$

Substituting for ( S_\infty ) and ( S_N ):

$160 - 20 \frac{1 - \left(\frac{7}{8}\right)^N}{1 - \frac{7}{8}} < 0.5$

This simplifies to:

$160 - 160(1 - \left(\frac{7}{8}\right)^N) < 0.5 \Rightarrow 160 \left(\frac{7}{8}\right)^N < 0.5$

Dividing by 160:

$\left(\frac{7}{8}\right)^N < \frac{0.5}{160} \approx 0.003125$

Taking logarithms:

$N \log \left(\frac{7}{8}\right) < \log(0.003125) \Rightarrow N > \frac{\log(0.003125)}{\log \left(\frac{7}{8}\right)}$

Calculating gives: ( N \approx 44.42 ). Thus, the smallest integer value is ( N = 45 ).